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Question 1: Find the zeroes of the quadratic polynomial $x^2 - 2x - 8$ and verify the relationship between the zeroes and the coefficients.
Solution: $x^2 - 2x - 8 = (x-4)(x+2)$. Zeroes are $4, -2$.
Sum $= 4 + (-2) = 2 = -(-2)/1 = -b/a$.
Product $= 4(-2) = -8 = -8/1 = c/a$. Verified. -
Question 2: Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively: $\frac{1}{4}, -1$.
Solution: Polynomial is $k[x^2 - (\text{sum})x + \text{product}]$.
$k[x^2 - \frac{1}{4}x - 1]$. Let $k=4$, then $4x^2 - x - 4$. -
Question 3: If $\alpha$ and $\beta$ are the zeroes of the quadratic polynomial $f(x) = x^2 - 5x + 4$, find the value of $\frac{1}{\alpha} + \frac{1}{\beta} - 2\alpha\beta$.
Solution: $\alpha+\beta = 5, \alpha\beta = 4$.
$\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha+\beta}{\alpha\beta} = \frac{5}{4}$.
Value $= \frac{5}{4} - 2(4) = \frac{5}{4} - 8 = \frac{5-32}{4} = -\frac{27}{4}$. -
Question 4: Find the zeroes of the quadratic polynomial $4\sqrt{3}x^2 + 5x - 2\sqrt{3}$.
Solution: $4\sqrt{3}x^2 + 8x - 3x - 2\sqrt{3} = 4x(\sqrt{3}x + 2) - \sqrt{3}(\sqrt{3}x + 2) = (4x - \sqrt{3})(\sqrt{3}x + 2)$.
Zeroes are $\frac{\sqrt{3}}{4}$ and $-\frac{2}{\sqrt{3}}$. -
Question 5: If one zero of the polynomial $(a^2 + 9)x^2 + 13x + 6a$ is reciprocal of the other, find the value of $a$.
Solution: Let zeroes be $\alpha, 1/\alpha$. Product $= 1$.
Product $= \frac{c}{a} = \frac{6a}{a^2+9} = 1 \Rightarrow a^2 - 6a + 9 = 0 \Rightarrow (a-3)^2 = 0 \Rightarrow a=3$. -
Question 6: If $\alpha$ and $\beta$ are zeroes of the quadratic polynomial $p(x) = x^2 - (k+6)x + 2(2k-1)$, find the value of $k$ if $\alpha + \beta = \frac{1}{2}\alpha\beta$.
Solution: $\alpha+\beta = k+6$, $\alpha\beta = 2(2k-1)$.
$k+6 = \frac{1}{2}[2(2k-1)] \Rightarrow k+6 = 2k-1 \Rightarrow k = 7$. -
Question 7: If the sum of the squares of zeroes of the polynomial $x^2 - 8x + k$ is 40, find the value of $k$.
Solution: $\alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta = 40$.
$(8)^2 - 2k = 40 \Rightarrow 64 - 2k = 40 \Rightarrow 2k = 24 \Rightarrow k = 12$. -
Question 8: Find a quadratic polynomial whose zeroes are $\frac{3+\sqrt{5}}{5}$ and $\frac{3-\sqrt{5}}{5}$.
Solution: Sum $= \frac{3+\sqrt{5} + 3-\sqrt{5}}{5} = \frac{6}{5}$.
Product $= \frac{9-5}{25} = \frac{4}{25}$.
Poly: $k[x^2 - \frac{6}{5}x + \frac{4}{25}]$. Let $k=25$, then $25x^2 - 30x + 4$. -
Question 9: If $\alpha$ and $\beta$ are the zeroes of the quadratic polynomial $f(x) = x^2 - p(x+1) - c$, show that $(\alpha+1)(\beta+1) = 1-c$.
Solution: $f(x) = x^2 - px - (p+c)$. $\alpha+\beta = p, \alpha\beta = -(p+c)$.
$(\alpha+1)(\beta+1) = \alpha\beta + \alpha + \beta + 1 = -(p+c) + p + 1 = -p - c + p + 1 = 1-c$. -
Question 10: If the zeroes of the polynomial $x^2 + px + q$ are double in value to the zeroes of $2x^2 - 5x - 3$, find the values of $p$ and $q$.
Solution: Zeroes of $2x^2 - 5x - 3$ are $3, -1/2$.
New zeroes are $6, -1$.
Sum $= 5 = -p \Rightarrow p = -5$. Product $= -6 = q \Rightarrow q = -6$.